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Find the area of the minor segment of a circle of radius $28 cm,$ when the angle of the corresponding sector is $45 °$ .

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112 c m^{2}

110 c m^{2}

11 c m^{2}

12 c m^{2}

answer is A.

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Detailed Solution

Given that, radius of minor segment of a circle is 28 cm and angle at the center is

45 °

.
We know that, area of a sector an angle at the center of the circle is equal to

π r 2 θ 360 °

.
Area of the triangle made by the chord of the circle which makes an angle at the center of the circle is equal to

12 r 2

.
Let

A_{1}

be the area of the sector.
Substitute 28 for r and 45° for

θ

π r 2 θ 360 °

to find the area of the sector.

A 1 = π 282 × 45 ° 360 °

⇒ 784 π 8

⇒

π

..............(1)
Let

A_{2}

be the area of the triangle.
Substitute 28 for r and 45° for

θ in 12 r 2 sin θ

to find the area of the triangle.

A 2 = 12282 sin 45 °

⇒ 12 × 784 × 12

⇒

196 ..........(2)
Let

A

be the area of the minor segment.
Subtract equation (2) from (1) to find the area of the minor segment.

A = A 1 − A 2

⇒ A = 98 π − 196

⇒

A = 98 × 22 7 − 196

⇒

A = 308

-

196

⇒

A = 112
Therefore, the area of the minor segment with corresponding sector that makes 45° at the center of a circle of radius 28 cm is 112

c m^{2}

.
Hence the correct option is 1.

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